$8.0575 \times 10^{-2}\text{ kg}$ of Glauber's salt is dissolved in water to obtain $1\text{ dm}^3$ of a solution of density $1077.2\text{ kg m}^{-3}$. Calculate the molarity, molality and mole fraction of $\text{Na}_2\text{SO}_4$ in the solution.

$0.25\text{ M},\; 0.2508\text{ m},\; 4.49 \times 10^{-3}$

The correct answer is Option (1) → $0.25\text{ M},\; 0.2508\text{ m},\; 4.49 \times 10^{-3}$ ##

Mass of Glauber's salt $= 8.0575 \times 10^{-2}\text{ kg}$

$= 8.0575 \times 10^{-2} \times 10^3\text{ g}$

$= 80.575\text{ g}$

Molecular mass of Glauber's salt $(\text{Na}_2\text{SO}_4 \cdot 10\text{H}_2\text{O}) = 322$

$\text{Number of moles of Glauber's salt} = \frac{80.575}{322} = 0.25$

$\text{Mass of solution per dm}^3 = 1077.2\text{ kg m}^{-3}$

$= 1077.2 \times 10^3\text{ g m}^{-3}$

$= 1077.2 \times 10^3 \times 10^{-3}\text{ g dm}^{-3}$

$= 1077.2\text{ g}$

$\text{Mass of water} = 1077.2 - 80.575 = 996.625\text{ g}$

$\text{Molarity} = \frac{0.25}{1\text{ dm}^3} = 0.25\text{ M}$

$\text{Molality} = \frac{0.25 \times 1000}{996.625} = 0.2508\text{ mol/kg}$

$\text{Mole fraction} = \frac{\text{Molarity}}{\text{Molarity} + \frac{\text{Mass of water}}{\text{Molecular mass of water}}}$

$= \frac{0.25}{0.25 + \frac{996.625}{18}} = 4.49 \times 10^{-3}$